Inverse tangents in a cyclic order 
If
  $$\theta= \tan^{-1}\left(\frac{a(a+b+c)}{bc}\right)+\tan^{-1}\left(\frac{b(a+b+c)}{ac}\right)+\tan^{-1}\left(\frac{c(a+b+c)}{ab}\right)$$then find $\tan\theta$

I tried to use these as sides of a triangle and use their properties, but other than that I am clueless. I cannot think of a substitution either. The answer happens to be zero. Any help is appreciated. Thanks in advance!!
 A: 
If
  $$\theta= \tan^{-1}\left(\color{red}{\sqrt{\frac{a(a+b+c)}{bc}}}\right)+\tan^{-1}\left(\color{red}{\sqrt{\frac{b(a+b+c)}{ac}}}\right)+\tan^{-1}\left(\color{red}{\sqrt{\frac{c(a+b+c)}{ab}}}\right)$$then $$\tan\theta=\color{red}{0}$$

Proof:
$$\tan^{-1}\left( \sqrt {\frac {x (x+y+z)}{yz}} \right)+\tan^{-1}\left( \sqrt {\frac {y (x+y+z)}{zx}} \right)+\tan^{-1}\left( \sqrt {\frac {z (x+y+z)}{xy}} \right)$$
Let  $ x $ - distance from the vertex $A $   of the triangle $ABC$ to the points of contact of the inscribed circle.
$y, z $ -  similary
Then: $x+y+z=p, yz= r\cdot r_a $ , where $ r_a $-    excircle radius.
$$\frac {x}{r}=\frac {p}{r_a}= ctg\left ( \frac {\alpha}{2} \right)$$
Then:  $$\alpha+\beta+\gamma =\pi $$
$$\tan^{-1}\left (ctg \left (\frac {\alpha}{2}\right)\right )+\tan^{-1}\left (ctg \left (\frac {\beta}{2}\right)\right )+\tan^{-1}\left (ctg \left (\frac {\gamma}{2}\right)\right )=\pi$$
Addition:

If
  $$\theta= \tan^{-1}\left(\frac{a(a+b+c)}{bc}\right)+\tan^{-1}\left(\frac{b(a+b+c)}{ac}\right)+\tan^{-1}\left(\frac{c(a+b+c)}{ab}\right)$$then find $\tan\theta$

Use: $$\tan(u+v+w)=\frac{\tan(u+v)+\tan w}{1-\tan(u+v)\cdot\tan w}=\frac{\tan u+\tan v+ \tan w-\tan u \cdot \tan v \cdot \tan w}{X}$$
Then $$\tan\theta=\frac{\left(\frac{a(a+b+c)}{bc}\right)+\left(\frac{b(a+b+c)}{ac}\right)+\left(\frac{c(a+b+c)}{ab}\right)-\left(\frac{a(a+b+c)}{bc}\right)\cdot\left(\frac{b(a+b+c)}{ac}\right)\cdot\left(\frac{c(a+b+c)}{ab}\right)}{X}$$
A: Use $$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}:$$
$$\theta= \tan^{-1}\left(\frac{a(a+b+c)}{bc}\right)+\tan^{-1}\left(\frac{b(a+b+c)}{ac}\right)+\tan^{-1}\left(\frac{c(a+b+c)}{ab}\right)=$$
$$=\tan^{-1}\frac{(a+b+c)^2}{1-(abc)^2}$$
Then
$$\tan \theta=\frac{(a+b+c)^2}{1-(abc)^2}$$
A: Okay, I have it. Apparently, there is no 'elegant' way to do this. Just use the equation for $arctan x + arctan y + arctan z$, and after some really messy calculation, you get the answer as zero. If there is a different, neat answer, please let me know.
